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2020 AMC 12A Problems/Problem 24

Revision as of 11:48, 16 March 2020 by Siluweston (talk | contribs) (→‎Solution)

Problem 24

Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP=1$, $BP=\sqrt{3}$, and $CP=2$. What is $s$?

$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$

Solution 1

[asy] draw((0,0)--(4,5.65)--(8,0)--cycle); label("A", (4,5.65), N, p = fontsize(10pt)); label("C", (8,0), SE, p = fontsize(10pt)); label("B", (0,0), SW, p = fontsize(10pt)); label("P", (3.5,3.5), NW, p = fontsize(10pt)); draw((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE); draw((8,0)--(3.5,3.5)); label("2",(8,0)--(3.5,3.5), SW); draw((4,5.65)--(3.5,3.5)); label("1",(4,5.65)--(3.5,3.5), E); draw((8,0)--(4,5.65)--(11,5.65)--cycle); label("$P'$", (6,4.75), NE, p = fontsize(10pt)); draw((4,5.65)--(6,4.75)); label("1",(4,5.65)--(6,4.75), S); draw((8,0)--(6,4.75)); label("$\sqrt{3}$",(8,0)--(6,4.75), E); draw((3.5,3.5)--(6,4.75)); label("1", (3.5,3.5)--(6,4.75), SE); [/asy]

We begin by rotating $\triangle{ABC}$ by $60^{\circ}$ about $A$, such that in $\triangle{A'B'C'}$, $B' = C$. We see that $\triangle{APP'}$ is equilateral with side length $1$, meaning that $\angle APP' = 60^{\circ}$. We also see that $\triangle{CPP'}$ is a $30-60-90$ right triangle, meaning that $\angle CPP'= 60^{\circ}$. Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$. We can now use the law of cosines as following:

\[s^2 = (AP)^2 + (CP)^2 - 2(AP)(CP)\cos{\angle{APC}}\] \[s^2 = 1 + 4 - 2(1)(2)\cos{120^{\circ}}\] \[s^2 = 5 - 4(-\frac{1}{2})\] \[s = \sqrt{5 + 2}\]

giving us that $s = \boxed{\textbf{(B) } \sqrt{7}}$. ~ciceronii

Solution 2 (Intuition)

[asy] unitsize(1inch); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label("A", (0,0), SW, p = fontsize(10pt)); label("C", (2.646,0), SE, p = fontsize(10pt)); label("B", (1.323,2.291), N, p = fontsize(10pt)); label("P'", (0.756,0.655), NW, p = fontsize(10pt)); label("1", (2.174,0.164), N, p = fontsize(10pt)); label("1", (1.228,0.491), N, p = fontsize(10pt)); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]

Suppose that triangle $ABC$ had three segments of length $2$, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments).

Take this equilateral triangle to have side length $1$. The portions of each segment outside this triangle (in red) have length $1$. Take $P'$ to be the intersection of the segments emanating from $A$ and $C$. By Law of Cosines, \[BP' = \sqrt{1 + 1 - 2\cos{120^\circ}} = \sqrt{3}.\] So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines \[s = \sqrt{4 + 1 - 4\cos{120^\circ}} = \boxed{\textbf{(B)} \sqrt{7}}.\]

~ hnkevin42

Video Solution

https://www.youtube.com/watch?v=mUW4zcrRL54

See Also

2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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